Question

Write the expression in standard form.

6/(4-13i)

Answer 1

a) -(24/185)- (78/185)i
b) (24/185)+(78/185)i
c)-(24/185)+(78/185)i
d) (24/185)i(78/185)i

Answer 2

It looks like these are all complex numbers. that need to be 'realized'. The trick is to do something similar to rationalising the denominator. We know that (a+ib)(a-ib) gives a^2+b^2 (a real number).
So all you do to each is multiply the numerator and denominator by the conjugate complex number
For 6/(4-13i) the conjugate of 4 - 13i is 4 +13i.
So multiply top and bottom by this.
6(4+13i)/(4^2 + 13^2) = (24 + 78i)/185 which is B.

Answer 3

i don't understand how you got 185 as the denominator or the other two numbers for the numerator by multiplying by the conjugate complex number...

Answer 4

For the denominator: Expand out (4-13i)(4+13i) This is a difference of two squares 4^2 - (13i)^2 but since i^2 = -1, by definition, then it becomes 4^2 - (-13^2) = 16 + 169 = 185. You always ADD the squares when multiplying a complex number with its conjugate.
The numerator is 6(4+13i) = 6x4 + 6x13i = 24 + 78i

Answer 5

Thank you so much! @mww